From AgBr precipitate:
0.75 g AgBr (M = 188) equals 0.00399 mol Br, so the compound has one Br atom.
From combustion of 1.0 g:
1.32 g CO₂ (M = 44) equals 0.03 mol C.
The molar mass is found from 0.53 g sample:
M = (0.53 × 188)/(0.75 × 1) = 133 g/mol.
From 1.0 g sample producing 0.03 mol CO₂
We have (1.0/133) × x = 0.03, giving x = 4.
Thus the molecular formula is C4H5Br with M = 48 + 5 + 80 = 133.
Percentage of hydrogen = (5/133) × 100 = 3.76%, which rounds to 4%.